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Two particles and gravity; non-constant acceleration

  1. Jul 4, 2008 #1

    I wanted to find some formulas for describing the motion of two particles.
    It seems, though, that it's harder than I originally thought.
    The situation:
    there are two particles (m1 and m2), and they are in rest somehow. Due to their mass, each particle attracts the other with a gravitational force [tex]F_g = \frac{G m_1 m_2}{d^2}[/tex], where G is the gravitational constant, d is the distance between the particles, and m1 and m2 are the masses of the particles.
    Combining this with: [tex]F=m \cdot a[/tex]
    Gives me: [tex]a_1 = \frac{G m_2}{d^2}[/tex]
    This is already useful, of course, but I'd rather have a formula that depends on time instead of the distance between the particles.
    So what I'd like to know: is there a formula for a(t)? (and/or v(t))

    I guess the formula would need to exist of at least G, the initial distance between the particles and their mass, but I really don't know how it would look like.
    I know that a(t) is not linear, and neither is v(t).
    Also, I know that acceleration is the integral of velocity with respect to time, but since I have no t in the formula for acceleration, I have no idea how to do this integral.

    (The motion of the particles is one-dimensional; they don't rotate around each other, they just move towards each other due to the gravitational forces they apply on each other.)

    I hope someone can help me out :biggrin:
  2. jcsd
  3. Jul 4, 2008 #2
    Hey Acronim,
    Yup you're quite right. You can get a(t) by a little calculus trick.Here:
    In order to avoid confusion let us say the distance between the bodies is "x" at a particular time.
    put a=vdv/dx = Gm2/x2....................(i)
    Integrate and find v(x). Now put v=dx/dt to get x(t)..............(ii)
    Once you have x(t) you can differentiate to find v(t) or a(t)
    Hope this solves your doubt.
  4. Jul 4, 2008 #3


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  5. Jul 4, 2008 #4
    Thanks for the fast (and rather useful) response!
    I'm not quite sure if I understand it completely, but I'll try to work it out. I'm no professor, you know (didn't even start at university yet):tongue2:... But I'll try to get as far as I can.
    What I don't quite understand is what exactly to integrate...
    You wrote:
    [tex]a= v \frac{dv}{dx}=\frac{Gm_2}{x^2}[/tex]
    Should I integrate only [tex]\frac{Gm_2}{x^2}[/tex]
    [tex]\int \frac{Gm_2}{x^2} dx=-\frac{Gm_2}{x}+c[/tex]
    Or should I integrate something else?

    I have quickly looked through that page, and also tried to find some more info about the two-body problem, but for as far as I could see, the two-body problem is about two body circulating around each other in some way. I don't know how I could possibly apply that if the bodies are not circulating around each other, but simply approaching each other like an apple approaches the earth in a straight line due to gravity.
  6. Jul 4, 2008 #5
    Considering that the bodies have no velocity at t=0. ok, That is a nice question
    basically by the simetry of the problem, you can consider a single axe, say x.

    Considering that 1 particle is still(to simplify), you have [tex]a_x=\frac{K}{x^2}[/tex] K is a constant. And the problem is there. That is called a differential equation. You have that the second derivative of x(acceleration) is equal to a constant times 1/x^2.

    That differential equation has no analytical solution. Basically that means that you can't derivate/integrate/math-tricks to solve it in order to x(t). So, no, you can't just get the equation of movement of that particle. A computer can, however, make a computational solution(an numerical approximation: with the two bodies it can too). If you wanna know more about it, ask here, and i post you a python code with an how to, or just pm me to talk msn.
  7. Jul 4, 2008 #6


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    It looks like I didn't read the original pos properly. The OP only wants to decribe the equations of motions of two masses originally at rest with each other, which effectively reduced the problem to 1 dimension. I think aniketp answered it pretty well. Here's what she/he meant. You have by the law of gravitation:

    [tex]a_1(x) = \frac{Gm_2}{x^2}[/tex]
    By the chain rule, [tex]a_1 = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v\frac{dv}{dx}[/tex]

    This is separable and solvable:

    [tex]\int a_1 dx = \int v dx[/tex]
    [tex]Gm_2 \int \frac{1}{x^2} dx = \int v dv[/tex]

    Once you get the expression for v in terms of x, note that v=dx/dt, which effectively allows you to solve by separation of variables to get x(t). The constants of integration can be obtained by substituting the initial conditions as you have given.
  8. Jul 4, 2008 #7

    D H

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    You are forgetting that the two particles are accelerating toward each other.
    Last edited: Jul 4, 2008
  9. Jul 4, 2008 #8


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    Which two other particles? There's only m1 and m2 here.
  10. Jul 4, 2008 #9

    D H

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    Yes, there are two particles, that was a typo. Fixed.

    Both accelerating toward each other. Post #4 and #6 deal with the acceleration of m1 with respect to inertial, not m1 with respect to m2.
  11. Jul 5, 2008 #10


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    I'm afraid I'm not following you here. What do you mean by respect to "inertial"? You have v1, v2 and these can be expressed as functions of x.
  12. Jul 5, 2008 #11

    D H

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    You said

    a_1 &= \frac {Gm2}{x^2} \\
    a_1 &= \frac {dv}{dt} = \frac {dv}{dx}\, \frac{dx}{dt}

    which is incorrect. One problem is the sign on the acceleration (first equation). The second equation is a bigger problem because it ignores the motion of mass 2 toward mass 1.

    Newton's law of gravitation applies to inertial frames. Your a1 is the magnitude of the acceleration of mass 1 with respect to some inertial frame. This is a mouthful, so many say "with respect to inertial" for short. The problem is that this is not the second time derivative of the distance between the two objects, which is what your second equation says.

    One way to reach the correct solution is to define x as the distance between the two masses, v as the time derivative of x, and a as the time derivative of v. From Newton's law of gravitation,

    [tex]a = -\,\frac {G(m_1+m_2)}{x^2}[/tex]

    Using the chain rule and separating as you did leads to the correct solution.
  13. Jul 5, 2008 #12


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    I think there is some confusion here. Apparently x is defined here as the separation distance between m1 and m2 at any one time, so v=dx/dt can't be correct. I was misled by the way in which x is used more commonly in other contexts. But if that is the problem then how can the chain rule and separation of variables work here be correct?
  14. Jul 5, 2008 #13
    Yeah that's the one you have to integrate. It's actually quite simple. You know that
    a=dv/dt .Using chain rule, a=dv/dx*dx/dt = vdv/dx . You may also try definite integration to get rid of the constant.
  15. Jul 5, 2008 #14

    D H

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    v=dx/dt is correct for the simple reason that this is the definition of v.
    Last edited: Jul 5, 2008
  16. Jul 5, 2008 #15

    D H

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    Except you are ignoring the motion of the other mass, and in doing so you get the wrong answer.
  17. Jul 5, 2008 #16
    The answer we will get is the relative velocity as a function of time, won't it?
  18. Jul 5, 2008 #17

    D H

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    Here are two ways to arrive at a description of the time evolution of the separation distance.

    Non-inertial frame
    Choose one of the masses -- say m1 -- as the origin of a reference frame. The goal is to describe the location of the other mass (m2) as a function of time. Denote the x-hat axis of our frame as pointing toward mass m2 and denote x as the distance between the two masses. The equation of motion in this accelerating frame is

    [tex]a \equiv \frac {d^2 x \hat x}{dt^2}
    = \frac {F_{\text{net}}}{m_2} - \ddot R[/tex]

    where F_net is the net force acting on mass m2 and R-double-dot is the frame acceleration. By Newton's law of gravitation and Newton's third law,

    F_{\text{net}} &= -\,\frac{Gm_1m_2}{x^2} \hat x \\
    \ddot R &= - \, \frac {F_{\text{net}}}{m_1} = \frac{Gm_2}{x^2} \hat x

    While the reference frame is accelerating, it is not rotating. The unit vector x-hat thusly is constant. Combining the above results,

    [tex]a =
    -\,\frac{Gm_1}{x^2} - \frac{Gm_2}{x^2} = -\,\frac{G(m_1+m_2)}{x^2}[/tex]

    Define [itex]v[/itex] as the first time derivative of [itex]x[/itex]. Then

    [tex]a = \frac {dv}{dt} = \frac{dv}{dx}\,\frac {dx}{dt} = v\frac{dv}{dx}[/tex]

    Combining with the previous result,

    [tex] v\frac{dv}{dx} = -\,\frac{G(m_1+m_2)}{x^2}[/tex]

    This is a separable and easily differential equation, giving v as a function of x. Integrating velocity with respect to time will yield the desired time desciption of the displacement vector.

    inertial frame
    Let [itex]\mathbf x_1(t)[/itex] and [itex]\mathbf x_2(t)[/tex] be the positions of masses m1 and m2 as functions of time from the perspective of some inertial frame. The goal is to describe the distance between the two masses as a function of time. By Newton's law of gravitation and Newton's second law,

    \frac{d^2\mathbf x_1}{dt} =
    \frac {G m_2}{||\mathbf x_2 - \mathbf x_1||^3} (\mathbf x_2 - \mathbf x_1) \\
    \frac{d^2\mathbf x_2}{dt} =
    \frac {G m_1}{||\mathbf x_1 - \mathbf x_2||^3} (\mathbf x_1 - \mathbf x_2)

    Taking the difference between these two acceleration vectors yields the second time derivative of the displacement vector from mass m1 to mass m2:

    [tex]\frac{d^2}{dt}(\mathbf x_2 - \mathbf x_1) =
    \frac {G m_1}{||\mathbf x_1 - \mathbf x_2||^3} (\mathbf x_1 - \mathbf x_2)
    - \frac {G m_2}{||\mathbf x_2 - \mathbf x_1||^3} (\mathbf x_2 - \mathbf x_1)
    = -\, \frac {G (m_1+m_2)}{||\mathbf x_1 - \mathbf x_2||^3}(\mathbf x_2 - \mathbf x_1)

    The desired goal is a description of the magnitude of the relative displacement vector. Define x as this magnitude and x-hat as the unit vector along the vector from mass m1 to mass m2. As the force will always be directed along this line, the unit vector x-hat is a constant unit vector. Thus

    [tex]\frac{d^2x}{dt}\hat x =
    -\, \frac {G (m_1+m_2)}{x^3}x \hat x

    Eliminating the unit vector x-hat and simplifying,

    [tex]a \equiv \frac{d^2x}{dt} =
    -\, \frac {G (m_1+m_2)}{x^2}

    The rest of the solution follows that of the non-inertial frame.
  19. Jul 5, 2008 #18
    Unfortunately not...a~1/(d^2) and d is not the same x as v = dx/dt is referring to. Therefore, the procedure used previously does not work.
  20. Jul 5, 2008 #19
    I know that my notation may be bad, since I leave out vector-notation and perhaps some other stuff I don't know about, but I thought it wasn't too important.
    I'll try to make a graphic representation ( = picture :tongue2:) of the situation.

    D H,
    I just don't really know why we should work with [tex]a=-\frac{G(m_1+m_2)}{x^2}[/tex]
    For as far as I understand, this is the difference between the acceleration of m1 and m2, so a2-a1, where a2 is "negative" and a1 is "positive". I know it works just as well as working with two accelerations seperately (It's easy to get a1 and a2 from a; same for v and x), or perhaps even easier.

    If it's easier to work with displacement x1 and x2, thus (x2-x1) instead of d or x as a distance between the particles, it's no problem of course.

    But if you've got a function for v as a function of x, how is it possible to integrate it with repect to time?

    Attached Files:

  21. Jul 5, 2008 #20

    D H

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    The distance between the two particles is x2-x1.

    Another way to look at it: Suppose the masses are not equal. Only using what you are calling a1 to describe the motion will lead to two different answers if you simply switch which particle you are calling m1 and which you are calling m2. The real behavior doesn't depend on the labels we apply to things.

    You should end up with a relationship [itex]v=f(x)[/tex]. In other words, [itex]dx/dt = f(x)[/itex]. You can solve for t as a function of x by integrating:

    [tex]t=\int \frac 1 {f(x)}\,dx = F(x)[/tex]

    If you want x as a function of t, you will have to find the inverse of this [itex]F(x)[/itex].
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